Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.2
Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and ‘an’ the nth term of the A.P.
Answer:
(i) Given: a = 7, d = 3, n = 8, find an
To find the 8th term (an) in the sequence, we use the formula:
an = a + (n – 1) * d
Substitute the given values:
an = 7 + (8 – 1) * 3
an = 7 + 7 * 3
an = 7 + 21 = 28
So, the 8th term (an) is 28.
(ii) Given: a = -18, n = 10, an = 0, find d
To find the common difference (d), we use the same formula:
an = a + (n – 1) * d
Substitute the given values:
0 = -18 + (10 – 1) * d
18 = 9 * d
d = 18 / 9 = 2
So, the common difference (d) is 2.
(iii) Given: d = -3, n = 18, an = -5, find a
To find the first term (a), use the formula:
an = a + (n – 1) * d
Substitute the given values:
-5 = a + (18 – 1) * (-3)
-5 = a + 17 * (-3)
-5 = a – 51
a = 51 – 5 = 46
So, the first term (a) is 46.
(iv) Given: a = -18.9, d = 2.5, an = 3.6, find n
To find the term number (n), use the formula:
an = a + (n – 1) * d
Substitute the given values:
3.6 = -18.9 + (n – 1) * 2.5
3.6 + 18.9 = (n – 1) * 2.5
22.5 = (n – 1) * 2.5
n – 1 = 22.5 / 2.5 = 9
n = 10
So, the term number (n) is 10.
(v) Given: a = 3.5, d = 0, n = 105, find an
To find the 105th term (an), use the formula:
an = a + (n – 1) * d
Substitute the given values:
an = 3.5 + (105 – 1) * 0 = 3.5
So, the 105th term (an) is 3.5.
Question 2
Choose the correct choice in the following and justify
(i) 30th term of the A.P: 10, 7, 4, …, is
(A) 97 (B) 77 (C) −77 (D.) −87
(ii) 11th term of the A.P. -3, -1/2, ,2 …. is
(A) 28 (B) 22 (C) – 38 (D) -48(1/2)
Answer:
(i) 30th term of the A.P: 10, 7, 4, …
Given:
First term (a) = 10
Common difference (d) = -3
We need to find the 30th term.
Formula:
aₙ = a + (n – 1) * d
For the 30th term:
a₃₀ = 10 + (30 – 1) * (-3)
a₃₀ = 10 + 29 * (-3)
a₃₀ = 10 – 87 = -77
So, the 30th term is -77.
The correct choice is (C) -77.
(ii) 11th term of the A.P. -3, -1/2, 2, …
Given:
First term (a) = -3
Common difference (d) = 5/2
We need to find the 11th term.
Formula:
aₙ = a + (n – 1) * d
For the 11th term:
a₁₁ = -3 + (11 – 1) * (5/2)
a₁₁ = -3 + 10 * (5/2)
a₁₁ = -3 + 25 = 22
So, the 11th term is 22.
The correct choice is (B) 22.
Final Answers:
(i) (C) -77
(ii) (B) 22
Question 3
In the following APs find the missing term in the boxes.
i) 2, _ , 26
ii) _ ,13, _ , 3
iii) 5, _ , , 9(1/2) v) , 38, , , _, -22
Answer:
i) 2, _, 26
We are given the first term (a) = 2 and the third term = 26. We need to find the missing second term.
Let the common difference be d.
We know that the nth term formula for an A.P. is:
aₙ = a + (n – 1) * d
For the third term:
a₃ = 26
Substitute the known values:
26 = 2 + (3 – 1) * d
26 = 2 + 2d
24 = 2d
d = 12
Now, we can find the second term:
a₂ = a + (2 – 1) * d = 2 + 1 * 12 = 2 + 12 = 14
So, the missing second term is 14.
ii) , 13, , 3
We are given the second term (a₂) = 13 and the fourth term (a₄) = 3. We need to find the missing first and third terms.
Let the first term be a and the common difference be d.
For the second term:
a₂ = a + (2 – 1) * d = a + d = 13
So, we have the equation:
a + d = 13 (Equation 1)
For the fourth term:
a₄ = a + (4 – 1) * d = a + 3d = 3
So, we have the equation:
a + 3d = 3 (Equation 2)
Now, solve the two equations:
From Equation 1:
a = 13 – d
Substitute into Equation 2:
(13 – d) + 3d = 3
13 + 2d = 3
2d = 3 – 13
2d = -10
d = -5
Now substitute d = -5 into Equation 1:
a + (-5) = 13
a = 13 + 5 = 18
So, the first term is 18.
Now, to find the third term:
a₃ = a + (3 – 1) * d = 18 + 2 * (-5) = 18 – 10 = 8
So, the missing terms are 18 (first term) and 8 (third term).
iii) 5, , , 9(1/2)
We are given the first term (a) = 5 and the fourth term (a₄) = 9.5. We need to find the missing second and third terms.
Let the common difference be d.
For the fourth term:
a₄ = a + (4 – 1) * d = a + 3d = 9.5
Substitute the known value for a:
5 + 3d = 9.5
3d = 9.5 – 5
3d = 4.5
d = 4.5 / 3 = 1.5
Now, we can find the second and third terms:
Second term:
a₂ = a + (2 – 1) * d = 5 + 1 * 1.5 = 5 + 1.5 = 6.5
Third term:
a₃ = a + (3 – 1) * d = 5 + 2 * 1.5 = 5 + 3 = 8
So, the missing terms are 6.5 (second term) and 8 (third term).
v) , 38, , , , -22
We are given the second term (a₂) = 38 and the sixth term (a₆) = -22. We need to find the missing first, third, fourth, and fifth terms.
Let the first term be a and the common difference be d.
For the second term:
a₂ = a + (2 – 1) * d = a + d = 38
So, we have the equation:
a + d = 38 (Equation 1)
For the sixth term:
a₆ = a + (6 – 1) * d = a + 5d = -22
So, we have the equation:
a + 5d = -22 (Equation 2)
Now, solve the two equations:
From Equation 1:
a = 38 – d
Substitute into Equation 2:
(38 – d) + 5d = -22
38 + 4d = -22
4d = -22 – 38
4d = -60
d = -60 / 4 = -15
Now substitute d = -15 into Equation 1:
a + (-15) = 38
a = 38 + 15 = 53
Now, we can find the missing terms:
First term: a = 53
Third term: a₃ = a + (3 – 1) * d = 53 + 2 * (-15) = 53 – 30 = 23
Fourth term: a₄ = a + (4 – 1) * d = 53 + 3 * (-15) = 53 – 45 = 8
Fifth term: a₅ = a + (5 – 1) * d = 53 + 4 * (-15) = 53 – 60 = -7
So, the missing terms are 53 (first term), 23 (third term), 8 (fourth term), and -7 (fifth term).
Summary of Missing Terms:
2, 14, 26
18, 13, 8, 3
5, 6.5, 8, 9.5
53, 38, 23, 8, -7, -22
Question 4
Which term of the A.P. 3, 8, 13, 18, … is 78?
Answer:
Formula: aₙ = a + (n – 1) * d
Where:
a = first term = 3
d = common difference = 8 – 3 = 5
aₙ = nth term (which is 78 in this case)
Substitute the values into the formula: 78 = 3 + (n – 1) * 5
Now, solve for n:
78 – 3 = (n – 1) * 5
75 = (n – 1) * 5
n – 1 = 75 / 5
n – 1 = 15
n = 16
So, the 16th term of the A.P. is 78.
Answer: The 16th term is 78.
Question 5
Find the number of terms in each of the following A.P.
(i) 7, 13, 19, …, 205
(ii) 18,15(1/2), 13,…., -47
Answer:
(i) For the A.P. 7, 13, 19, …, 205:
The first term is 7, the common difference is 6 (since 13 – 7 = 6), and the last term is 205.
Using the formula for the nth term of an A.P : aₙ = a + (n – 1) * d
Substitute the values:
205 = 7 + (n – 1) * 6
Now, solve for n:
205 – 7 = (n – 1) * 6
198 = (n – 1) * 6
n – 1 = 198 / 6
n – 1 = 33
n = 34
So, there are 34 terms in the A.P.
(ii) For the A.P. 18, 15(1/2), 13, …, -47:
The first term is 18, the common difference is -2.5 (since 15.5 – 18 = -2.5), and the last term is -47.
Using the same formula:
aₙ = a + (n – 1) * d
Substitute the values:
-47 = 18 + (n – 1) * (-2.5)
Now, solve for n:
-47 – 18 = (n – 1) * (-2.5)
-65 = (n – 1) * (-2.5)
n – 1 = -65 / -2.5
n – 1 = 26
n = 27
So, there are 27 terms in this A.P.
The number of terms in the first A.P. is 34.
The number of terms in the second A.P. is 27.
Question 6
Check whether -150 is a term of the A.P. 11, 8, 5, 2, …
Answer:
To check whether -150 is a term of the A.P. 11, 8, 5, 2, …, we need to find out if -150 can be represented as a term of this sequence.
The formula for the nth term of an A.P. is:
aₙ = a + (n – 1) * d
Where:
a = first term = 11
d = common difference = 8 – 11 = -3
aₙ = -150 (the term we are checking for)
Now, substitute the known values into the formula:
-150 = 11 + (n – 1) * (-3)
Simplifying the equation:
-150 – 11 = (n – 1) * (-3)
-161 = (n – 1) * (-3)
Now, divide both sides by -3:
n – 1 = 161 / 3
n – 1 = 53.67
n = 54.67
Since n is not a whole number (it’s a decimal), -150 cannot be a term of this A.P.
Conclusion: -150 is not a term of the A.P. 11, 8, 5, 2, ….
Question 7
Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.
Answer:
Given:
11th term = 38
16th term = 73
Step 1: Find the common difference (d)
We use this formula for any term in an A.P. (Arithmetic Progression):
nth term = first term + (n – 1) × common difference
For the 11th term: 38 = first term + (10 × common difference)
For the 16th term: 73 = first term + (15 × common difference)
Step 2: Subtract the first equation from the second
This helps us get rid of the first term and find the common difference:
(73 – 38) = (15d – 10d)
This simplifies to: 35 = 5d
Now, solve for d: d = 35 ÷ 5 = 7
So, the common difference (d) is 7.
Step 3: Find the first term
Now that we know d = 7, we can find the first term. We use the first equation again:
38 = first term + (10 × 7)
38 = first term + 70
So, the first term is: first term = 38 – 70 = -32
Step 4: Find the 31st term
Now, use the formula again to find the 31st term:
31st term = first term + (30 × common difference)
Substitute the values we know: 31st term = -32 + (30 × 7)
31st term = -32 + 210
31st term = 178
Final Answer:
The 31st term is 178.
Question 8
An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Answer:
The formula for the nth term of an A.P. is:
aₙ = a + (n – 1) * d
Where:
aₙ is the nth term,
a is the first term,
d is the common difference between the terms.
Step 1: Write the equations for the 3rd and 50th terms
For the 3rd term: a₃ = a + (3 – 1) * d = 12 This simplifies to: 12 = a + 2d (Equation 1)
For the 50th term: a₅₀ = a + (50 – 1) * d = 106 This simplifies to: 106 = a + 49d (Equation 2)
Step 2: Subtract the two equations to find the common difference d
Now, subtract Equation 1 from Equation 2: (a + 49d) – (a + 2d) = 106 – 12 Simplifying this gives: 47d = 94 Solving for d: d = 94 / 47 = 2
Step 3: Find the first term a
Now that we know d = 2, substitute this value into Equation 1: 12 = a + 2 * 2 12 = a + 4 a = 12 – 4 = 8 So, the first term is a = 8.
Step 4: Find the 29th term
To find the 29th term, use the formula: a₂₉ = a + (29 – 1) * d Substitute a = 8 and d = 2: a₂₉ = 8 + 28 * 2 a₂₉ = 8 + 56 = 64
Final Answer:
The 29th term of the arithmetic progression is 64.
Question 9
If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.
Answer:
We are given the following information about the arithmetic progression (A.P.):
The 3rd term is 4.
The 9th term is -8.
We need to find which term of this A.P. is zero.
Step 1: Use the formula for the nth term of an A.P.
The formula for the nth term of an A.P. is:
an = a + (n – 1) * d
Where:
an is the nth term,
a is the first term,
d is the common difference.
Step 2: Set up equations for the 3rd and 9th terms
For the 3rd term: a3 = a + (3 – 1) * d = 4 a + 2d = 4 (Equation 1)
For the 9th term: a9 = a + (9 – 1) * d = -8 a + 8d = -8 (Equation 2)
Step 3: Subtract the two equations to find d
Now, subtract Equation 1 from Equation 2:
(a + 8d) – (a + 2d) = -8 – 4 6d = -12 d = -12 / 6 = -2
Step 4: Find the first term a
Now that we know d = -2, substitute this value into Equation 1 to find a:
a + 2(-2) = 4 a – 4 = 4 a = 4 + 4 = 8
So, the first term is a = 8.
Step 5: Find the term that is zero
We need to find which term is zero. Set an = 0 and use the formula:
0 = a + (n – 1) * d Substitute a = 8 and d = -2:
0 = 8 + (n – 1) * (-2) 0 = 8 – 2(n – 1) 0 = 8 – 2n + 2 0 = 10 – 2n 2n = 10 n = 10 / 2 = 5
Final Answer: The 5th term of the A.P. is zero.
Question 10
If 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
Answer:
Given that the 17th term exceeds the 10th term by 7, we can write the equation:
a₁₇ = a₁₀ + 7
Using the formula for the nth term (aₙ = a + (n – 1) * d), we can express the 17th and 10th terms:
a₁₇ = a + 16d
a₁₀ = a + 9d
Now, substitute these into the equation:
a + 16d = (a + 9d) + 7
Simplifying the equation:
a + 16d = a + 9d + 7
Cancel out “a” from both sides:
16d = 9d + 7
Subtract 9d from both sides:
7d = 7
Solve for d:
d = 7 / 7 = 1
So, the common difference is 1.
Question 11
Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term?
Answer:
We are given:
a = 3 (first term)
d = 12 (common difference)
Find the 54th term (a₅₄):
a₅₄ = a + (54 – 1) * d = 3 + 53 * 12 = 3 + 636 = 639
Add 132 to the 54th term:
639 + 132 = 771
Find which term is 771:
We use the formula for the nth term:
aₙ = a + (n – 1) * d
Substitute aₙ = 771, a = 3, and d = 12:
771 = 3 + (n – 1) * 12
771 – 3 = (n – 1) * 12
768 = (n – 1) * 12
n – 1 = 768 / 12 = 64
n = 64 + 1 = 65
Final Answer:
The 65th term is 132 more than the 54th term.
Question 12
Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?
Answer:
we need to find the difference between their 1000th terms.
Formula for the nth term of an AP:
The formula for the nth term of an AP is: aₙ = a + (n – 1) * d
Where:
aₙ is the nth term,
a is the first term,
d is the common difference.
Find the 100th terms of both APs:
Let the first AP be denoted by a₁ and the second AP by a₂.
For the 100th term of the first AP: a₁₀₀ = a₁ + 99 * d
For the 100th term of the second AP: a₂₀₀ = a₂ + 99 * d
We are told that the difference between these terms is 100: |a₁₀₀ – a₂₀₀| = 100
This simplifies to: |a₁ – a₂| = 100
Find the 1000th terms of both APs:
For the 1000th term of the first AP: a₁₀₀₀ = a₁ + 999 * d
For the 1000th term of the second AP: a₂₀₀₀ = a₂ + 999 * d
The difference between the 1000th terms is: a₁₀₀₀ – a₂₀₀₀ = (a₁ + 999 * d) – (a₂ + 999 * d)
Simplifying: a₁₀₀₀ – a₂₀₀₀ = a₁ – a₂
Since we know that |a₁ – a₂| = 100, the difference between the 1000th terms is also 100.
Final Answer:
The difference between their 1000th terms is 100.
Question 13
How many three digit numbers are divisible by 7?
Answer:
The three-digit numbers divisible by 7 are: 105, 112, 119, …, 994.
These numbers form an arithmetic progression (AP) with:
First term = 105
Common difference = 7
We want to find the number of terms in this AP. The nth term is 994, so using the formula for the nth term of an AP:
aₙ = a + (n – 1) * d
Substitute the values: 994 = 105 + (n – 1) * 7
Now, subtract 105 from both sides: 994 – 105 = (n – 1) * 7
889 = (n – 1) * 7
Now, divide both sides by 7: n – 1 = 889 / 7 = 127
Finally, add 1 to both sides: n = 127 + 1 = 128
So, there are 128 three-digit numbers divisible by 7.
Question 14
How many multiples of 4 lie between 10 and 250?
Answer:
We are asked to find how many multiples of 4 lie between 10 and 250.
The first multiple of 4 greater than 10 is 12. The next multiple is 16, and so on.
Therefore, the multiples of 4 between 10 and 250 are: 12, 16, 20, 24, …, 248.
These numbers form an arithmetic progression (AP) with:
First term = 12
Common difference = 4
To find the number of terms, we need to find
𝑛
n such that the nth term of this AP is 248.
Using the formula for the nth term of an AP:
aₙ = a + (n – 1) * d
Substitute the known values:
248 = 12 + (n – 1) * 4
Now, subtract 12 from both sides:
248 – 12 = (n – 1) * 4
236 = (n – 1) * 4
Divide both sides by 4:
n – 1 = 236 / 4 = 59
Finally, add 1 to both sides:
n = 59 + 1 = 60
So, there are 60 multiples of 4 between 10 and 250.
Question 15
For what value of n, are the nth terms of two APs 63, 65, 67, and 3, 10, 17, … equal?
Answer:
63, 65, 67, …
a = 63
d = a2 − a1 = 65 − 63 = 2
nth term of this A.P. = an = a + (n − 1) d
an= 63 + (n − 1) 2 = 63 + 2n − 2
an = 61 + 2n … (i)
3, 10, 17, …
a = 3
d = a2 − a1 = 10 − 3 = 7
nth term of this A.P. = 3 + (n − 1) 7
an = 3 + 7n − 7
an = 7n − 4 … (ii)
It is given that, nth term of these A.P.s are equal to each other.
Equating both these equations, we obtain
61 + 2n = 7n − 4
61 + 4 = 5n
5n = 65
n = 13
Therefore, 13th terms of both these A.P.s are equal to each other
Question 16
Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12
Answer:
We are given:
The third term of the A.P. is 16.
The 7th term exceeds the 5th term by 12.
Step 1: Use the formula for the nth term of an A.P.
The formula for the nth term of an A.P. is: aₙ = a + (n – 1) * d
Where:
aₙ is the nth term,
a is the first term,
d is the common difference.
Step 2: Use the given information
The third term is 16: a₃ = a + (3 – 1) * d = a + 2d = 16 So, we get the equation: a + 2d = 16 (Equation 1)
The 7th term exceeds the 5th term by 12: a₇ – a₅ = 12 Using the formula for the nth term: (a + 6d) – (a + 4d) = 12 Simplifying: 2d = 12 So, d = 6
Step 3: Solve for the first term (a)
Now, substitute d = 6 into Equation 1: a + 2(6) = 16 a + 12 = 16 a = 16 – 12 = 4
Final Answer:
The A.P. is: 4, 10, 16, 22, …
Question 17
Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.
Answer:
We are given the A.P. as 3, 8, 13, …, 253.
The first term (a) is 3, the common difference (d) is 5, and the last term (l) is 253.
We need to find the 20th term from the last term.
Step 1: Find the total number of terms (n) in the A.P.
Use the formula for the nth term of an A.P.:
aₙ = a + (n – 1) * d
Substitute the known values (aₙ = 253, a = 3, and d = 5):
253 = 3 + (n – 1) * 5
Now, subtract 3 from both sides:
250 = (n – 1) * 5
Divide both sides by 5:
50 = n – 1
So, n = 51.
Thus, there are 51 terms in total.
Step 2: Find the 20th term from the last term.
The 20th term from the last term will be the (51 – 20 + 1) = 32nd term from the first term.
Now, use the formula for the nth term again to find the 32nd term:
a₃₂ = a + (32 – 1) * d a₃₂ = 3 + 31 * 5 a₃₂ = 3 + 155 a₃₂ = 158
Final Answer:
The 20th term from the last term is 158.
Question 18
The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.
Answer:
Given:
a₄ + a₈ = 24
a₆ + a₁₀ = 44
We have the equations:
a + 5d = 12
a + 7d = 22
Solving these:
Subtract equation 1 from equation 2: 2d = 10 d = 5
Now, substitute d = 5 into a + 5d = 12: a + 25 = 12 a = -13
First three terms:
a = -13
a₂ = a + d = -13 + 5 = -8
a₃ = a₂ + d = -8 + 5 = -3
So, the first three terms are -13, -8, and -3.
Question 19
Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?
Answer:
It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.
Therefore, the salaries of each year after 1995 are
5000, 5200, 5400, …
Here, a = 5000
d = 200
Let after nth year, his salary be Rs 7000.
Therefore, an = a + (n − 1) d
7000 = 5000 + (n − 1) 200
200(n − 1) = 2000
(n − 1) = 10
n = 11
Therefore, in 11th year, his salary will be Rs 7000.
Question 20
Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nth week, her week, her weekly savings become Rs 20.75, find n.
Answer:
It can be observed that Ramkali’s savings are in A.P. as every week her savings increase by Rs 1.75.
Therefore, the savings in each week are:
5, 6.75, 8.5, 10.25, …
Here, a = 5
d = 1.75
Let after nth week, her savings be Rs 20.75.
Therefore, an = a + (n − 1) d
20.75 = 5 + (n − 1) 1.75
1.75(n − 1) = 15.75
(n − 1) = 15.75 / 1.75
(n − 1) = 9
n = 10
Therefore, in the 10th week, her savings will be Rs 20.75.
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